Question

8. The accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [e=1.6×10⁻¹⁹C; Me=9.1×10⁻³¹kg]

Accepted Answer

Given: V = 2.5kV = 2.5×10³V, Me = 9.1×10⁻³¹kg, charge q of an electron = 1.6×10⁻¹⁹C.
Using ½Mev² = qV: v² = 2qV/Me = (2×1.6×10⁻¹⁹×2.5×10³)/(9.1×10⁻³¹) = 8.79×10¹⁴. v = √(8.79×10¹⁴) = 2.965×10⁷ ms⁻¹.

WAEC Physics 2013 Theory — Question 8

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